Since $S\cap x^{-1}S$ is more than half of $G$, so is $C(x)$. So by Lagrange's Theorem, $C(x) = G$, and $x$ is in the center of $G$. Thus $S$ is a subset of the center, and it is more than half of $G$. So the center must be all of $G$, that is $G$ is commutative. Once $G$ is commutative the problem is easy.

If f does not invert more than 3/4 of the elements of G, then the result is false. Take Q={+-1,+-i,+-j,+-k} the order 8 quaternion group, and let f(i)=-i and f(j)=-j (this determines f since i and j generate Q). Then f sends +-1, +-i, and +-j to their inverses (thats 6 out of 8, which is 3/4), but does not send k to its inverse.
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Anton GeraschenkoSep 30 '09 at 2:29

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An observation: since f\circ f must be the identity on more than half of the elements of G, it must be the identity.
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Anton GeraschenkoOct 1 '09 at 3:28

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I think the point of this whole 3/4 business is the following. If G_1 is the set of elements such that f(x) = x^{-1}, then if we look at left multiplication on G by an element of G_1, more than half the elements have to make back into G_1.

Combining this with what we know about f it should follow that any g \in G_1 commutes with more than 1/2 the elements of G, so if you say Langrange's thm enough times it should follow that G is abelian and G_1 generates G, which together imply the result.

(My girlfriend explained this to me.) After Anton's observation, it's sufficient to show that f = id if f fixes more than half of G. But the elements of G fixed by an automorphism form a group and this group has index less than 2 by assumption, hence is all of G.

I don't follow. f does not fix more than half of G, it sends more than 3/4 of the elements to their inverses. It looks like this proves that f\circ f is the identity because it is an automorphism that fixes more than half of G.
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Anton GeraschenkoOct 1 '09 at 14:30